Problem: $f(0)=4\,$, $~f\,^\prime(0)=2\,$, $f\,^{\prime\prime}(0)=1\,$, and $~f\,^{\prime\prime\prime}(0)=-3\,$. What are the first four nonzero terms of the Maclaurin series of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4+2x+{{x}^{2}}-3{{x}^{3}}$ (Choice B) B $4+2x+\frac{1}{2}{{x}^{2}}-{{x}^{3}}$ (Choice C) C $4+2x+\frac{1}{2}{{x}^{2}}+\frac{1}{2}{{x}^{3}}$ (Choice D) D $4+2x+\frac{1}{2}{{x}^{2}}-\frac{1}{2}{{x}^{3}}$ (Choice E) E $4x+2x^2+\frac{1}{2}{{x}^{3}}-\frac{1}{2}{{x}^{4}}$
Answer: We know the formula for a Taylor series centered at $~x=0~$ is $ f(0)+f\,^\prime(0)x+\frac{f\,^{\prime\prime}(0)}{2!}{{x}^{2}}+\frac{f\,^{\prime\prime\prime}(0)}{3!}{{x}^{3}}+...+\frac{{{f}^{(n)}}(0)}{n!}{{x}^{n}}+...\,$ So, if we substitute with what is given for the first four terms, we get the following. $ T_3(x)=4+2x+\frac{1}{2!}{{x}^{2}}-\frac{3}{3!}{{x}^{3}}$ This simplifies to $ T_3(x)=4+2x+\frac{1}{2}{{x}^{2}}-\frac{1}{2}{{x}^{3}}\,$.